Đáp án đúng: B
Giải chi tiết:\(\begin{array}{l}\int\limits_0^{\frac{\pi }{4}} {f'\left( x \right)\sin 2xdx} = \int\limits_0^{\frac{\pi }{4}} {\sin 2xd\left( {f\left( x \right)} \right)} = \left. {\left( {\sin 2x.f\left( x \right)} \right)} \right|_0^{\frac{\pi }{4}} - \int\limits_0^{\frac{\pi }{4}} {f\left( x \right)d\left( {\sin 2x} \right)} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( {\dfrac{\pi }{4}} \right) - 0 - 2\int\limits_0^{\frac{\pi }{4}} {f\left( x \right)\cos 2xdx} = - 2\int\limits_0^{\frac{\pi }{4}} {f\left( x \right)\cos 2xdx} = - \dfrac{\pi }{4}\\ \Rightarrow \int\limits_0^{\frac{\pi }{4}} {f\left( x \right)\cos 2xdx} = \dfrac{\pi }{8} \Rightarrow \int\limits_0^{\frac{\pi }{4}} {{f^2}\left( x \right)dx} - \int\limits_0^{\frac{\pi }{4}} {f\left( x \right)\cos 2xdx} = 0\\ \Leftrightarrow \int\limits_0^{\frac{\pi }{4}} {\left( {{f^2}\left( x \right) - f\left( x \right)\cos 2x} \right)dx} = 0 \Leftrightarrow {f^2}\left( x \right) - f\left( x \right)\cos 2x = 0 \Leftrightarrow \left[ \begin{array}{l}f\left( x \right) = 0\\f\left( x \right) = \cos 2x\end{array} \right.\end{array}\)
\(f\left( x \right) = 0\): Loại, do \(\int\limits_0^{\frac{\pi }{4}} {{f^2}\left( x \right)dx} = \dfrac{\pi }{8}\)
\(f\left( x \right) = \cos 2x \Rightarrow \)\(I = \int\limits_0^{\frac{\pi }{8}} {f\left( {2x} \right)dx = } \int\limits_0^{\frac{\pi }{8}} {\cos 4xdx} = \left. {\dfrac{1}{4}\sin 4x} \right|_0^{\frac{\pi }{8}} = \dfrac{1}{4}\).
Chọn: B
